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Edexcel A-level Mathematics (9MA0) · Moments
Mini-Lesson

Moments

This mini-lesson covers Topic 9 of Edexcel A-level Mathematics (9MA0): moments in static contexts. You will find the moment of a force (including a force acting at an angle), use the two conditions for the equilibrium of a rigid body, work with uniform and non-uniform rods on supports, and solve tilting problems. Take g = 9.8 m s⁻².

pivot F₁ d₁ F₂ d₂ anticlockwise clockwise in equilibrium: F₁d₁ = F₂d₂
Moment = force × perpendicular distance from the pivot, measured in N m. Balanced ⇒ total clockwise = total anticlockwise.

Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Throughout, take g = 9.8 m s−2 unless a question says otherwise. Press Start when you are ready.

Moments · definition

The moment of a force

A moment is the turning effect of a force about a point.

moment = F × dd is the PERPENDICULAR distance from the point to the line of action of F · units: N m
  • State the sense: clockwise or anticlockwise.
  • A force whose line of action passes through the point has zero moment about it (d = 0).
  • If the force acts at an angle θ to the rod, only the perpendicular component turns it: moment = F d sin θ.
Worked example

A force of 30 N acts perpendicular to a rod, 0.6 m from the pivot.

Moment = 30 × 0.6 = 18 N m.

Now suppose a 20 N force acts 3 m from the pivot at 40° to the rod:

Moment = F d sin θ = 20 × 3 × sin 40° = 60 × 0.6428 = 38.6 N m (3 s.f.)

Calculate

A simple moment

1A force of 30 N acts perpendicular to a rod at a distance of 0.6 m from the pivot. Find the moment about the pivot.
N m
Hint: Moment = F × d = 30 × 0.6.
Calculate

Force at an angle

2A force of 20 N acts at a point 3 m from the pivot, at an angle of 40° to the rod. Find the moment about the pivot, to 3 significant figures.
N m
Hint: Only the perpendicular component turns the rod: moment = F d sin θ = 20 × 3 × sin 40°.
Quick check

Zero moment

?What is the moment about a pivot of a force whose line of action passes through the pivot?
Moments · equilibrium

Equilibrium of a rigid body

A rigid body in equilibrium needs two conditions, not one:

  • Resultant force = 0 — so the forces balance vertically (and horizontally).
  • Total moment about any point = 0 — so total clockwise moment = total anticlockwise moment.

The weight of a uniform rod acts at its midpoint (its centre of mass). For a non-uniform rod, the centre of mass is somewhere else — and finding it is often the question.

A B C R_A R_C W = mg a UNIFORM rod's weight acts at its midpoint
Two unknown reactions? Take moments about one of them — its own moment is then zero, so it drops out of the equation.
Worked example — seesaw

A uniform beam is pivoted at its midpoint. A 30 N weight sits 1.5 m to the left of the pivot. Where must a 45 N weight sit on the right to balance it?

Anticlockwise = clockwise: 30 × 1.5 = 45 × d → 45 = 45d → d = 1 m from the pivot.

(The beam's own weight acts at the pivot, so it has zero moment about it and can be ignored here.)

The key tactic: if there are two unknown reactions, take moments about one of them. That reaction has zero moment about its own point of action, so it vanishes and you are left with one unknown.

Sort it

Uniform rod, equilibrium, or tilting?

Tap a statement, then tap where it belongs.

📏 True of a uniform rod

⚖️ Needed for equilibrium

↕️ About tilting

Quick check

Two conditions

?For a rigid body to be in equilibrium, which condition(s) must hold?
Moments · rods on supports

Uniform rods on two supports

Worked example

A uniform rod AB of length 4 m and mass 6 kg rests horizontally on supports at A and at C, where AC = 3 m.

Weight = 6 × 9.8 = 58.8 N, acting at the midpoint, 2 m from A.

Moments about A (this kills RA): RC × 3 = 58.8 × 2 → RC = 117.6 ÷ 3 = 39.2 N

Resolving vertically: RA + RC = 58.8 → RA = 58.8 − 39.2 = 19.6 N

Check by taking moments about C: RA × 3 = 58.8 × 1 → RA = 19.6 ✔

Always check. Taking moments about a second point is a free, independent check of both reactions — and the support nearer the centre of mass always carries more.

Calculate

Reaction at C

3A uniform rod AB of length 4 m and mass 6 kg rests on supports at A and at C, where AC = 3 m. Taking g = 9.8, find the reaction RC.
N
Hint: Weight = 6 × 9.8 = 58.8 N at the midpoint, 2 m from A. Moments about A: R_C × 3 = 58.8 × 2.
Calculate

Reaction at A

4For the same rod, find the reaction RA at A.
N
Hint: Resolve vertically: R_A + R_C = 58.8, and R_C = 39.2 N.
Moments · non-uniform rods

Non-uniform rods — finding the centre of mass

If a rod is not uniform, its weight does not act at the midpoint. Let the centre of mass be a distance d from one end and solve for d.

Worked example

A non-uniform rod AB of length 5 m and mass 8 kg rests on supports at A and B. The reaction at A is 30 N. Find the distance of the centre of mass from A.

Weight = 8 × 9.8 = 78.4 N. Resolving vertically: RA + RB = 78.4 → RB = 78.4 − 30 = 48.4 N.

Moments about A: 78.4 × d = RB × 5 = 48.4 × 5 = 242

d = 242 ÷ 78.4 = 3.09 m (3 s.f.) from A — past the midpoint, towards B, as you would expect since RB > RA.

Sense check: the centre of mass always lies nearer the support with the larger reaction. If your answer disagrees, re-check the moments.

Calculate

Centre of mass

5A non-uniform rod AB of length 5 m and mass 8 kg rests on supports at A and B. The reaction at A is 30 N. Taking g = 9.8, find the distance of the centre of mass from A, to 3 significant figures.
m
Hint: W = 8 × 9.8 = 78.4 N, so R_B = 78.4 − 30 = 48.4 N. Moments about A: 78.4 × d = 48.4 × 5.
Quick check

Where does the weight act?

?The weight of a uniform rod acts at which point?
Moments · tilting

Tilting: the reaction that vanishes

When a plank on two supports is on the point of tilting about one support, it is about to lift off the other one — so the reaction at that other support becomes zero. This single fact solves every tilting problem.

Worked example

A uniform plank AB of length 6 m and mass 20 kg rests on supports at C (1 m from A) and D (4 m from A). A child of mass 30 kg walks from A towards B. How far from A is the child when the plank is about to tilt about D?

About to tilt about D ⇒ RC = 0. The plank's weight (20g) acts at the midpoint, 3 m from A — that is 1 m to the left of D. Let the child be x m from A, so (x − 4) m to the right of D.

Moments about D: 30g × (x − 4) = 20g × 1

The g cancels: 30(x − 4) = 20 → x − 4 = 20 ÷ 30 = 0.667 → x = 4.67 m (3 s.f.) from A.

Take moments about the support it is turning about — then the zero reaction and the pivot reaction both disappear at once, leaving a single equation.

Calculate

Tilting point

6A uniform plank AB of length 6 m and mass 20 kg rests on supports at C (1 m from A) and D (4 m from A). A child of mass 30 kg walks from A towards B. Find the distance from A at which the plank is about to tilt about D, to 3 significant figures.
m
Hint: On the point of tilting about D, R_C = 0. Moments about D: 30g(x − 4) = 20g × 1 (the plank's weight is 1 m from D). The g cancels.
Match it

Term → its meaning

Tap an item on the left, then its partner on the right.

Item
Meaning
Quick check

About to tilt

?A plank resting on two supports is on the point of tilting about the support at D. What is true about the reaction at the other support, C?
Recap

The big ideas to know

Moment = F × perpendicular distance, in N m — state clockwise or anticlockwise

At an angle: moment = F d sin θ · a force through the point has zero moment

Equilibrium of a rigid body: resultant force = 0 and total moment about any point = 0

Uniform rod: the weight acts at the midpoint · non-uniform: solve for the centre of mass

Tactic: take moments about an unknown reaction to eliminate it

Tilting: about to tilt about one support ⇒ the reaction at the other support is zero

That is the whole of 9MA0 Topic 9 — Moments. Press Finish to see your score.

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